Question

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X+1/x=4 find x2+1/x2

Answer 1

Correct Question :-

$\sf If\;\; x + \dfrac{1}{x} = 4 \;then\;find\; x^{2} + \dfrac{1}{x^{2}}$

Given :-

$\sf \bullet \;\; x + \dfrac{1}{x} = 4$

To Find :-

$\sf \bullet \;\; x^{2} + \dfrac{1}{x^{2}}$

Solution :-

~Here, we’re given the value of x + 1/x which is 4 and in order to find the values of x² + 1/x² we will do the squaring on both the sides.  

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Let’s solve !  

$\sf \implies x + \dfrac{1}{x} = 4$

$\sf \maltese \;\; ( Squaring\;on\;both\;sides)$

$\sf \implies \bigg\{ x + \dfrac{1}{x} \bigg\}^{2} = \{ 4 \}^{2}$

$\sf \maltese \;\; ( a+b)^{2} = (a)^{2} + (b)^{2} + (2ab)$  

$\sf \implies \{ x \}^{2} + \bigg\{ \dfrac{1}{x} \bigg\}^{2} + \bigg\{ 2 \times x \times \dfrac{1}{x} \bigg\} = 16$

$\sf \implies x^{2} + \dfrac{1}{x^{2}} + \dfrac{2x}{x} = 16$

 

$\sf \implies x^{2} + \dfrac{1}{x^{2}} + 2 = 16$

 

$\sf \implies x^{2} + \dfrac{1}{x^{2}} = 16-2$

$\sf {\dag} \;\; x^{2} + \dfrac{1}{x^{2}} = 14$

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Hence ,  

• The answer is 14  

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More to know :-

• a² – b² = (a – b)(a + b)
• (a + b)² = a² + 2ab + b²
• a² + b² = (a + b)² – 2ab
• (a – b)² = a² – 2ab + b²
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ca

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Answer 2

$x + \frac{1}{x} = 2 \\ \\ \\ \\ (x + \frac{1}{2} {)}^{2} = {2}^{2} \\ \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 4 \\ \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 2 \\ \\ \\ (x - \frac{1}{x} {)}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 2 - 2 = 0 \\ {x}^{2} - \frac{1}{ {x}^{2} } = 0$